Acta Mathematica Sinica, English Series July, 2002, Vol.18, No.3, pp. 597–604

Chen’s Theorem with Small Primes Ying Chun CAI Department of Mathematics, Shanghai University, Shanghai 200436, P. R. China

Abstract

Let N be a sufficiently large even integer. In this paper it is proved that the equation p ≤ N 0.95 ,

N = p + P2 ,

is solvable, where p denotes a prime and P2 denotes an almost prime with at most two prime factors. Keywords Chen’s theorem, Sieve, Mean value theorem MR(2000) Subject Classification 11N36

1

Introduction

In 1966 Chen [1] made considerable progress in the research of the binary Goldbach conjecture, he [2] proved the well-known Chen’s Theorem: Let N be a sufficiently large even integer, then the equation N = p + P2 ,

(1.1)

is solvable, where p is a prime and P2 is an almost prime with at most two prime factors. In fact, Chen’s Theorem can be stated in a more exact quantitative form: Let S(N ) denote the number of solutions to equation (1.1), then S(N ) ≥ where C(N ) =


 p>2

1−

0.67C(N )N , log2 N 1 (p − 1)2




p|N,p>2

p−1 . p−2

The constant 0.67 was improved by Halberstam [3], Chen [4,5], and the best one 0.8285 is due to the present author and Prof. Lu [6]. In this paper we shall obtain Chen’s Theorem with small primes. Theorem

Let S(N, θ) denote the number of solutions to the equation N = p + P2 ,

p ≤ N θ.

(1.2)

Received April 13, 1999, Accepted December 18, 2000 Project supported by The National Natural Science Foundation of China (Grant no. 10171060, 19801021) and by MCSEC

Cai Y. C.

598

Then, for θ = 0.95, S(N, θ) >

2

0.01C(N )N θ . log2 N

Some Lemmas

Let A denote a finite set of integers, P denote an infinite set of primes, P denote the set of primes that do not belong to P. Let z ≥ 2, put P (z) =




p,

1,

a∈A,(a,P (z))=1

p
Ad = {a|a ∈ A, a ≡ 0(modd)}, Lemma 1 [7]



S(A, P, z) =

P(q) = {p|p ∈ P, (p, q) = 1}.

If A1 ) |Ad | = A2 )

 z1 ≤p
ω(d) X + rd , µ(d) = 0, (d, P) = 1; d   log z2 ω(p) 1 = log , z2 > z1 ≥ 2, +O p log z1 log z1

where ω(d) is a multiplicative function, 0 ≤ ω(p) < p, X > 1 is independent of d, then    1 S(A, P, z) ≥ XV (z) f (s) + O − RD , 1 log 3 D    1 S(A, P, z) ≤ XV (z) F (s) + O + RD 1 log 3 D where s=

log D , log z



RD =

|rd |,

d
 1 , log z  −1
 1 ω(p) 1− 1− C(ω) = , p p p V (z) = C(ω)

e−γ log z





1+O

here γ denoting the Euler constant, f (s) and F (s) being determined by the following differentialdifference equation:  γ  F (s) = 2e , f (s) = 0, 0 < s ≤ 2, s  (sF (s)) = f (s − 1), (sf (s)) = F (s − 1), s ≥ 2. Lemma 2 [8] F (s) =

2eγ , 0 < s ≤ 3; s

Chen’s Theorem with Small Primes



599





s−1 2eγ log(t − 1) 1+ dt , 3 ≤ s ≤ 5; s t 2 2eγ log(s − 1) f (s) = , 2 ≤ s ≤ 4; s   s−1 2eγ dt t−1 log(u − 1) f (s) = log(s − 1) + du , 4 ≤ s ≤ 6. s t 2 u 3

F (s) =

Lemma 3 [8] (Bombieri) For any given constant A > 0, there exists a constant B = B(A) > 0 1 such that, for D = x 2 log−B x,



 x Liy

 max max 1− ,

ϕ(d) (l,d)=1 y≤x logA x p≤y d≤D

p≡l(modd)

where Lix =

x

dt . 2 log t

Lemma 4 [9]

Let g(n) be an arithmetical function such that  g 2 (n) logc x, n

n≤x

for some c > 0. For (al, q) = 1, define 

1 H(z, h, a, q, l) = 1− ϕ(q) z≤ap


 Li

z+h a

 − Li

 z  a

.

ap≡l(q)

Then, for any given constant A > 0, there exists a constant B = B(A, c) > 0 such that, for 3 1 5θ − 3 < θ ≤ 1, y = xθ , 0 ≤ β < , λ = θ − , D = xλ log−B x, 5 2 2



 

y

max max xmax g(a)H(z, h, a, d, l)

. (l,d)=1 h≤y 2 ≤z≤x logA x

d≤D a≤xβ ,(a,d)=1 3

Weighted Sieve Method

Let N be a sufficiently large even integer and put   A = a a = N − p, p ≤ N θ , θ = 0.95,

(3.1)

P = {p| (p, N ) = 1}.

(3.2)

Lemma 5 S(N, θ) ≥

 a∈A 1

(a,P (N 11 ))=1

  10 1 1 1 − ρ1 (a) − ρ2 (a) − ρ3 (a) + O(N 11 ), 2 2

Cai Y. C.

600

where



ρ1 (a) = N 

ρ2 (a) =  ρ3 (a) = Proof

1,

p|a,(p,N )=1 1 1 11 ≤p
1

1, a = p1 p2 p3 , N 11 ≤ p1 < N 3.29 ≤ p2 < p3 , (a, N ) = 1, 0, otherwise; 1

1, a = p1 p2 p3 , N 3.29 ≤ p1 < p2 < p3 , (a, N ) = 1, 0, otherwise.

Let v1 (a) =



1, v2 (a) =



 1, λ(a) =

pm |a

p|a

1, v2 (a) ≤ 2, 0, v2 (a) > 2.

Then S(N, θ) ≥





λ(a) =

a∈A 1

λ(a)+O(v1 (N )) =

a∈A,(a,N )=1 1

(a,P (N 11 ))=1



µ2 (a)λ(a)+O(N 11 ). 10

a∈A,(a,N )=1 1

(a,P (N 11 ))=1

(a,P (N 11 ))=1

On the other hand, 

 a∈A 1

1 1 1 − ρ1 (a) − ρ2 (a) − ρ3 (a) 2 2

(a,P (N 11 ))=1

=

 a∈A,(a,N )=1 1



  10 1 1 µ2 (a) 1 − ρ1 (a) − ρ2 (a) − ρ3 (a) + O(N 11 ). 2 2

(a,P (N 11 ))=1

For µ2 (a) = 1, (a, P (N 11 )) = 1, (a, N ) = 1 we have: 1

1) v2 (a) ≤ 2, λ(a) = 1 ≥ 1 − 12 ρ1 (a) − 12 ρ2 (a) − ρ3 (a); 2) v2 (a) ≥ 3, if ρ1 (a) ≥ 2, then λ(a) = 0 ≥ 1 − 12 ρ1 (a) − 12 ρ2 (a) − ρ3 (a). If ρ1 (a) = 1, then v1 (a) = v2 (a) = 3, and ρ2 (a) = 1, hence λ(a) = 0 = 1 − 12 ρ1 (a) − 12 ρ2 (a) − ρ3 (a). If ρ1 (a) = 0, then ρ3 (a) = 1, and λ(a) = 0 = 1 − 12 ρ1 (a) − 12 ρ2 (a) − ρ3 (a). Combining the above arguments we complete the proof of Lemma 5. 4

Proof of the Theorem

In this section, the sets A and P are defined by (3.1) and (3.2) respectively, and X = LiN θ ∼

Nθ . log N θ

For (d, N ) = 1, rd = π(N θ ; d, N ) −

d LiN θ , ω(d) = , µ(d) = 0, (d, N ) = 1. ϕ(d) ϕ(d)

Chen’s Theorem with Small Primes

601

By Lemma 5 we have 10 1 1 S(N, θ) ≥ S − S1 − S2 − S3 + O(N 11 ), 2   2 1 1, S1 = S(Ap , P, N 11 ), S= 1 1 N 11 ≤p
a∈A,(a,N )=1 1

(a,P (N 11 ))=1

(p,N )=1



S2 =

(4.1)



ρ2 (a), S3 =

a∈A,(a,N )=1 1 (a,P (N 11 ))=1

ρ3 (a).

a∈A,(a,N )=1 1 (a,P (N 11 ))=1

1) A lower bound for S. θ

Let D = RD and

N2 logB N

with B = B(5) > 0. By Lemma 3 we get





Nθ LiN θ

 Liy

θ

= max max π(y; d, l) − ,

π(N ; d, N ) − ϕ(d) ≤ ϕ(d) y≤N θ (l,d)=1 log5 N d≤D d≤D

(4.2)

 −1
 −1
  −1
 1 1 ω(p) 1 1 1− 1− 1− 1− 1− = C(ω) = p p p p−1 p p p|N (p,N )=1 
p
 p
p(p − 2) (p − 1)2
p(p − 2) · = = 2 p−1 (p − 1)2 p − 1 p(p − 2) p>2 (p − 1)2 p|N (p,N )=1 p|N,p>2 

 1 p−1 = 2C(N ). (4.3) 1− =2 2 (p − 1) p−2 p>2 p|N,p>2

By Lemmas 1 and 2, (4.2), (4.3) and some routine arguments, we get that   5.5θ−2 5.5θ − 1 log(s − 1) C(N )N θ log ds S ≥ 8(1 + o(1)) 2 log(5.5θ − 1) + s s+1 θ log2 N 2 ≥ 12.9897

C(N )N θ . log2 N

(4.4)

2) An upper bound for S1 . θ

1

1

with B = B(5) > 0. For N 11 ≤ p < N 3.29 , by Lemma 1 we get that   log p 1 Nθ −γ 11 S(Ap , P, N ) ≤ 22(1 + o(1))C(N )e F 5.5θ − 11 + RD (p), (4.5) log N pθ log2 N

Let D =

N2 logB N

where



RD (p) =

d< D p ,d|P (N

By Lemma 3 we have  1 1 N 11 ≤p
(p,N )=1

|rdp |. 1 11

)





1 1 N 11 ≤p
11 ) d< D p ,d|P (N

RD (p) =

(p,N )=1

1

θ

π(N θ ; dp, N ) − LiN

ϕ(dp)



Nθ Liy

≤ max max π(y; d, l) − . ϕ(d) (l,d)=1 y≤N θ log5 N d≤D 

(4.6)

Cai Y. C.

602

By (4.5) and (4.6), the prime number theorem and summation by parts we get that    log p Nθ 1 S1 ≤ 22(1 + o(1))C(N )e−γ × F 5.5θ − 11 p log N θ log2 N 1 1 N 11 ≤p
(p,N )=1

1   N 3.29 log u Nθ 1 × F 5.5θ − 11 du = 22(1 + o(1))C(N )e 1 u log u log N θ log2 N N 11     5.5θ−2 log(s − 1) C(N )N θ 11θ − 2 (5.5θ − 1)(5.5θ − 1 − s) ≤ 8(1 + o(1)) log + log ds 3.29θ − 2 s s+1 θ2 log2 N 2



−γ

≤ 18.7347

C(N )N θ . log2 N

(4.7)

3) Upper bounds for S2 , S3 . By the definition of ρ2 (a) we have  S2 =



1 1 N 11 ≤p1
( )

(p1 p2 ,N )=1

N p1

1 2

p2


=



1 1 N 11 ≤p1
( )

(p1 p2 ,N )=1

N p1

1 2



≤

1

p=N −p1 p2 p3 ,p≤N θ p2


1 1 N 11 ≤p1
(p1 p2 ,N )=1

1

a∈A,a=p1 p2 p3

( pN1 )

1 2

N −N θ p1 p2

1

p=N −p1 p2 p3


= Σ.

(4.8)

Consider the sets     12

1 1 N

11 3.29 E = e e = p1 p2 , N ≤ p1 < N ≤ p2 < , (p1 p2 , N ) = 1 ,

p1   L = l l = N − ep, e ∈ E, N − N θ < ep < N, (p, N ) = 1 . We have 



|E| ≤ N

1 11

≤p1
1 3.29

N p1

 12

2

< N 3,

1

2

N 3 < e ≤ N 3 , e ∈ E, 1

l < N θ , l ∈ L. 2

The number of elements in the set L which are less than N 3 does not exceed N 3 , Σ does not exceed the number of primes in L. Therefore, 2

1

Σ ≤ S(L, P, z) + O(N 3 ), z ≤ N 3 . Now we apply Lemma 1 to estimate the upper bound of S(L, P, z). For the set L,     N  N − Nθ X= − Li , Li e e e∈E

(4.9)

Chen’s Theorem with Small Primes

603

ω(d) =

d , ϕ(d)

µ(d) = 0, (d, N ) = 1,

1

N θ− 2 1 D= , B = B(5) > 0, z = D 2 . B log N Then, by Lemma 1, we get that S(L, P, z) ≤ where

8 X (1 + o(1))C(N ) + R1 + R2 , 2θ − 1 log N

(4.10)

 





   

1

 

(Li(E 1 − ) − Li(E )) R1 = 2 1  ,

 ϕ(d)

E1
e∈E

(e,d)=1 ep3 ≡N (d) E1 = R2 =

N − Nθ , e  d≤D,(d,N )=1

1

N E2 = , e  1 (Li(E2 ) − Li(E1 )). ϕ(d) e∈E,(e,d)>1



2

In view of N 3 ≤ e < N 3 for e ∈ E, letting g(a) =

e=a,e∈E

1, then we have

 



    

1

R1 = (Li(E2 ) − Li(E1 )) g(a)  1−

 

,

1 ϕ(d) 2 E1
N 3
ap3 ≡N (d) (a,d)=1



R2 =

d≤D,(d,N )=1

1 ϕ(d)





g(a)

1.

E1
2 1 N 3
It is easy to show that g(a) ≤ 1. Now R2

 d≤D

1 ϕ(d)

N log N

 2 1 N 3


d≤D

N

10 11

1 ϕ(d)

 1 N a ϕ(d) d≤D

 1



 1

m|d,m≥N 11

1 N log N m

2 a
(a,d)=m

 1

N 11 ≤m≤D

m|d,m≥N 11

N a

log2 N.

 1 1 mϕ(m) D ϕ(d) d≤ m

(4.11)

By Lemma 4 we get that R1

Nθ . log5 N

(4.12)

By Huxley’s prime number theorem in short intervals and integeration by parts, we have X = (1 + o(1))

 (Li(E2 ) − Li(E1 )) e∈E

Cai Y. C.

604



= (1 + o(1))N θ N

1 11

= (1 + o(1))N

≤p1
1

N 11

= (1 + o(1))

Nθ log N



≤p2 <

N p1 p2

p1

( Nt ) 2 dt du 1 t log t N 3.29 u log u log   log 2.29 − 3.29 s+1 ds. s 1

1 3.29

N

θ

1 3.29

1  N  12 p1 p2 log

10

2.29

N ut

(4.13)

By (4.8)–(4.13) we get that S2 ≤ 8(1 + o(1))

θ

C(N )N (2θ − 1) log2 N



 log 2.29 −

10

3.29 t+1



t

2.29

dt ≤ 6.8086

C(N )N θ . log2 N

(4.14)

Similarly, we have S3 ≤ 8(1 + o(1))



C(N )N θ (2θ − 1) log N

1

N 3.29 ≤p1
≤ 8(1 + o(1))

θ

C(N )N (2θ − 1) log2 N



2

2.29

1  N  12 p1 p2 log

N p1 p2

p1

C(N )N θ log(t − 1) . dt ≤ 0.1564 t log2 N

(4.15)

4) Proof of the Theorem. By (4.1), (4.4), (4.7), (4.14) and (4.15) we get   0.01C(N )N θ 18.7347 6.8085 C(N )N θ > . S(N, θ) ≥ 12.9897 − − − 0.1564 2 2 2 log N log2 N The theorem is proved. References [1] Chen, J. R.: On the representation of a large even integer as the sum of a prime and the product of at most two primes. Kexue Tongbao, 17, 385–386 (1966) [2] Chen, J. R.: On the representation of a large even integer as the sum of a prime and the product of at most two primes. Sci. Sin., 16, 157–176 (1973) [3] Halberstam, H.: A proof of Chen’s Theorem. Asterisque, 24–25, 281–293 (1975) [4] Chen, J. R.: On the representation of a large even integer as the sum of a prime and the product of at most two primes. Sci. Sin., 21, 421–430 (1978) [5] Chen, J. R.: On the representation of a large even integer as the sum of a prime and the product of at most two primes. Sci. Sin., 21, 477–494 (1978) (in Chinese) [6] Lu, M. G., Cai, Y. C.: On Chen’s Theorem, to appear [7] Iwaniec, H.: Rosser’s sieve, Recent Progress in Analytic Number Theory II, London: Academic Press, 203–230 (1981) [8] Pan, C. D., Pan, C. B.: Goldbach Conjecture, Beijing: Science Press, 175–176 (1992) [9] Wu, J.: Theoremes generalises de Bombieri-Vinogradov dans les petits applications, intervalles. Quart. J. Math., (Oxford), 44, 109–128 (1993)